如何使用 express 获取异步方法的输出值作为输入和显示

Question

我对node和express有点陌生

我正在尝试构建一个 API 消耗外部 API 和 returns 结果作为我显示内容的输入

如何使用快速发送方法获取异步方法的输出值作为输入和显示

查看下面我的快递代码

const express = require("express");
var AsyncMethod = require("./src/AsyncMethod");//works perfectly
const app = express();
const port = 8000;

app.get("/url", (req, res) => {
var output = "";
  var biller = new AsyncMethod();
  try {
    output = await (biller.run());
    .then(//need help here
      console.log("inside Main")
    res.send(output))
  } catch (ex) {
    console.log("An error occurred");
    res.send("An error occurred");
    console.log(ex);
  }
});

Answer 1

您需要使用 .then() 方法和 return 回调的响应等待异步方法执行完成。

使用Async/Await

const express = require("express");
var AsyncMethod = require("./src/AsyncMethod");//works perfectly
const app = express();
const port = 8000;

app.get("/url", async (req, res) => {
    var output = "";
    var biller = new AsyncMethod();
    try {
        output = await biller.run();
        res.send(output)
    } catch (ex) {
      console.log("An error occurred");
      res.send("An error occurred");
      console.log(ex);
    }

});

使用promise

const express = require("express");
var AsyncMethod = require("./src/AsyncMethod");//works perfectly
const app = express();
const port = 8000;

app.get("/url", (req, res) => {
    var biller = new AsyncMethod();
    biller.run().then(function(output){
      res.send(output)
    }).catch(function(ex){
      console.log("An error occurred");
      res.send("An error occurred");
      console.log(ex);
    })
});

Answer 2

我查看了您在评论中发布 link 的 github 项目

.run 需要 return 一个 Promise 如果你想 await 它

那么 - 这是如何做到这一点...另外 5 行标记为 //+++

var BaseSample = require('./BaseSample');

var GetBillers = function(){
    //inherit
    BaseSample.call(this);
}

GetBillers.prototype.run = function(){
  return new Promise((resolve, reject) => {  //+++
    this.billpayment.get_billers(function(err, res){
        if(err) {
            //error executing request
            console.log("Error calling get billers "+JSON.stringify(err));
            reject(err); //+++
        } else{
            //check if it was successful
            var statusCode = res.statusCode;

            if(statusCode === 200) {//request was successful

                var billerArray = JSON.parse(res.body).billers;
                var firstBiller = billerArray[0];

                var billerId = firstBiller.billerid;
                var billername = firstBiller.billername;
                console.log(billerId+" "+billername);
                resolve(billerId+" "+billername); //+++
            }
            else{//it was not successful for a reason
                console.log("FAILED: "+statusCode);
                reject(statusCode); //+++
            }
        }
    });
  }); //+++
}

module.exports = GetBillers;

哦，你贴的代码应该有点像

const express = require("express");
var AsyncMethod = require("./src/AsyncMethod");//works perfectly
const app = express();
const port = 8000;

app.get("/url", (req, res) => {
  var biller = new AsyncMethod();
  try {
    const output = await biller.run();
    console.log("inside Main")
    res.send(output))
  } catch (ex) {
    console.log("An error occurred");
    res.send("An error occurred");
    console.log(ex);
  }
});

如何使用 express 获取异步方法的输出值作为输入和显示

How do I get the output value of an async method as input and display using express

javascript

nodes

node.js

express

typescript